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\begin{document}
%
% paper title
% can use linebreaks \\ within to get better formatting as desired
\title{Deformación de un cilindro mediante calor}
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\author{\textbf{Autores:} Alejandro Magnorsky, Andrés Mata Suárez, Mariano Merchante \\[5px]
        Instituto Tecnológico de Buenos Aires}
        
% note the % following the last \IEEEmembership and also \thanks - 
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\renewcommand{\abstractname}{Resumen}


\renewcommand{\IEEEkeywordsname}{Palabras clave}
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\begin{abstract}
\boldmath
En este artículo se expone un método numérico para calcular la deformación de un cilindro debido a su temperatura.
Primero se obtiene la función de temperatura del cilindro en función del radio y del tiempo, resolviendo una ecuación
diferencial a derivadas parciales. Luego, se encuentra la deformación del cilindro realizando una integral por medio del método de los trapecios. 

\end{abstract}
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% Note that keywords are not normally used for peerreview papers.
\begin{IEEEkeywords}
Cilindro; Temperatura; Deformación; Diferencias finitas.
\end{IEEEkeywords}






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\vspace{1cm}


\section{Introducción}

La necesidad de calcular las deformaciones de un cilindro en base a su temperatura es de mucha utilidad. Desde pistones
en pequeños motores hasta maquinarias industriales, pasando por circuitos eléctricos, se vuelve de trascendental importancia poder preveer su comportamiento en base a su calentamiento o enfriamiento. \\

La sección \ref{section:development} muestra principalmente el desarrollo de la ecuación diferencial a derivadas parciales que describe a la función de temperatura del cilindro. Luego se expone la fórmula para obtener la deformación del cilindro en función del tiempo. \\

Luego, en la seccion \ref{section:results} se muestran los resultados de los programas realizados junto con gráficos que permiten visualizar con mayor precisión la función de temperatura y la de deformación. \\

Por último, la sección \ref{section:conclusions} ofrece conclusiones realizadas una vez obtenidos los resultados.


\vspace{1cm}
%-------------------------------------------------------------------------
\section{Desarrollo}
\label{section:development}
\vspace{0.5cm}
\subsection{Cálculo de la temperatura del cilindro}

La temperatura de un cilindro uniforme puede modelarse como una función $u(r,t)$, donde $r$ es la coordenada radial desde el eje del cilindro y $t$ es el tiempo. Dicha función debe satisfacer la siguiente ecuación diferencial:
\begin{equation}
\frac{\partial^{2} u}{\partial r^{2}} + \frac{1}{r} \frac{\partial u}{\partial r} = \frac{1}{4K} \frac{\partial u}{\partial t}
\label{equation:diff}
\end{equation}
para $\frac{1}{2} < r < 1$, $0< t < 10$ y $K = 0.1$. \\

Las condiciones de contorno, para $0 \leq t \leq 10$, son: \\
\begin{equation}
u(\frac{1}{2}, t) = t
\label{equation:contour1}
\end{equation}
\begin{equation}
u(1, t) = 100 + 40t
\label{equation:contour2}
\end{equation}


Además, la condición inicial, para $\frac{1}{2} \leq r \leq 1$, es:
\begin{equation}
u(r, 0) = 200(r-0.5)
\label{equation:initial}
\end{equation}


El objetivo es, entonces, encontrar una aproximación $v_{k}^{m} \approx u(\frac{1}{2} + k\Delta r, m\Delta t)$, donde $k = 0,1,\dots,n$ 
siendo $n = \frac{1-\frac{1}{2}}{\Delta r}$ y $m = 0,1,\dots,l$ con $l = \frac{10}{\Delta t}$. \\

Para ello, se utilizan las siguientes diferencias finitas:
\begin{itemize}
\item Centradas \\
\begin{equation}
\frac{\partial^{2} u}{\partial r^{2}} = \frac{u(r+\Delta r, t) - 2 u(r,t) + u(r-\Delta r,t)}{\Delta r^{2}} + O(\Delta r^{2})
\end{equation}
\begin{equation}
\frac{\partial u}{\partial r} = \frac{u(r+\Delta r, t) - u(r-\Delta r,t)}{2\Delta r} + O(\Delta r^{2})
\end{equation}
\item Progresiva \\
\begin{equation}
\frac{\partial u}{\partial t} = \frac{u(r, t + \Delta t) - u(r,t)}{\Delta t} + O(\Delta t)
\end{equation}
\end{itemize}

Escribiendo las aproximaciones de las diferencias centradas en la ecuación \eqref{equation:diff}, se obtiene el siguiente esquema: \\
\begin{equation}
\frac{v_{k+1}^{m} - 2v_{k}^{m} + v_{k-1}^{m}}{\Delta r^{2}} + \frac{1}{\frac{1}{2}+k\Delta r} \frac{v_{k+1}^{m}-v_{k-1}^{m}}{2\Delta r} = \frac{1}{4K} \frac{v_{k}^{m+1} - v_{k}^{m}}{\Delta t}
\end{equation}
\begin{equation}
v_{k}^{m+1} = v_{k}^{m} + \frac{4K\Delta t}{\Delta r^{2}} (v_{k+1}^{m} - 2v_{k}^{m} + v_{k-1}^{m}) + \frac{4K\Delta t}{(\frac{1}{2}+k\Delta r)2\Delta r} (v_{k+1}^{m}-v_{k-1}^{m})
\label{equation:diffapprox}
\end{equation}

Las condiciones de contorno y la inicial se deducen de las ecuaciones \eqref{equation:contour1}, \eqref{equation:contour2} y \eqref{equation:initial} :
\begin{equation}
v_{0}^{m} = m\Delta t 
\label{equation:vcontour1}
\end{equation}
\begin{equation}
v_{n}^{m} = 100 + 40m\Delta t 
\label{equation:vcontour2}
\end{equation}
\begin{equation}
v_{k}^{0} = 200(\frac{1}{2}+k\Delta r - 0.5) 
\label{equation:vinitial}
\end{equation}

Las diferencias finitas, utilizadas para construir el esquema de la ecuación \eqref{equation:diffapprox}, tienen errores de $O(\Delta r^{2})$ y $O(\Delta t)$. Por lo tanto, el esquema tiene $O(\Delta r^{2} + \Delta t)$, es decir, orden (2,1). \\

Para analizar la estabilidad, se aplica la Transformada de Fourier de Tiempo Discreto en ambos miembros del esquema de la ecuación \eqref{equation:diffapprox}, junto a las propiedades de desplazamiento en el tiempo y linealidad \citep{mathews}, obteniendo:

\begin{equation}
V_{k}^{m+1} = V_{k}^{m} + \frac{4K\Delta t}{\Delta r^{2}} (V_{k}^{m} e^{i\omega}- 2V_{k}^{m} + V_{k}^{m} e^{-i\omega}) + \frac{4K\Delta t}{r 2\Delta r} (V_{k}^{m} e^{i\omega} - V_{k}^{m} e^{-i\omega})
\end{equation}

Despejando $V_{k}^{m}$:

\begin{equation}
V_{k}^{m+1} = (1 + \frac{4K\Delta t}{\Delta r^{2}} (e^{i\omega} + e^{-i\omega}) - 2\frac{4K\Delta t}{\Delta r^{2}} + \frac{4K\Delta t}{r 2\Delta r} (e^{i\omega} - e^{-i\omega})) V_{k}^{m}
\end{equation}
\begin{equation}
V_{k}^{m+1} = (1 + 2 \frac{4K\Delta t}{\Delta r^{2}}(\cos \omega - 1) + i 2 \frac{4K\Delta t}{r 2\Delta r} \sin \omega) V_{k}^{m}
\end{equation}

Se define $\rho = 1 + 2 \frac{4K\Delta t}{\Delta r^{2}}(\cos \omega - 1) + i 2 \frac{4K\Delta t}{r 2\Delta r} \sin \omega$. Entonces:

\begin{equation}
V_{k}^{m+1} = \rho V_{k}^{m}
\end{equation}
\begin{equation}
V_{k}^{m+1} = \rho^{m+1} V_{k}^{0}
\end{equation}

Por lo tanto, $|\rho| < 1$ para que el método sea estable. Para que eso se cumpla, como mínimo, $|Im(\rho)| < 1$:
\begin{equation}
|2 \frac{4K\Delta t}{r 2\Delta r} \sin \omega| <1 \Rightarrow \frac{4K\Delta t}{r 2\Delta r} < \frac{1}{2} \Rightarrow \Delta t < \frac{r\Delta r}{4 K}
\end{equation}

Considerando los casos en que $r = \frac{1}{2}$ y $r = 1$, se obtiene que el caso más restrictivo es que $\Delta t < \frac{\Delta r}{8 K}$. \\

En el listado \ref{lst:finiteDifferences}, se detalla la implementación de un programa que, dados $\Delta r$ y $\Delta t$, utiliza las condiciones de contorno y la inicial (ecuaciones \eqref{equation:vcontour1}, \eqref{equation:vcontour2} y \eqref{equation:vinitial}) para armar parte de la matriz $v$ que representa a la función $v_{k}^{m}$, donde $k$ es la columna y $m$ es la fila. Para completar el resto, usa el esquema de la ecuación \eqref{equation:diffapprox}. \\



\subsection{Cálculo de la deformación del cilindro}

La deformación del cilindro, $d$, es proporcional a la temperatura media del mismo. Por lo tanto, se calcula como:
\begin{equation}
d(t) = \alpha \int_{\frac{1}{2}}^{1} u(r,t)r dr
\label{equation:deformation}
\end{equation}
donde $\alpha = 10.7$. \\

Para el cálculo de la ecuación \eqref{equation:deformation} se utiliza el método de los trapecios. \\

En el listado \ref{lst:stableDef}, se detalla la implementación del cálculo de la deformación para cualquier valor de $t$. En particular, para $tIndex = \frac{10}{\Delta t} + 1$, se obtiene la deformación para el régimen permanente, es decir, para $t = 10$.


\vspace{1cm}
%-------------------------------------------------------------------------
\section{Resultados}
\label{section:results}
\vspace{0.5cm}

Para realizar todos los cálculos se utiliza $\Delta r = 0.1$ y $\Delta t = 0.01$. \\

En la figura \ref{figure:graphC} pueden verse los resultados de las curvas de temperatura para $ r = 0.6,0.7,0.8 $ y $0.9$. Estas curvas demuestran que el cilindro tiene temperaturas mayores según la distancia al eje central, y además crece con una tendencia lineal a medida que transcurre el tiempo. Esto se debe a las condiciones de contorno en $r=\frac{1}{2}$ y $r=1$, ya que son crecientes en el tiempo. \\

Además, en la figura \ref{figure:mesh} se visualiza el campo $u(r,t)$. Como puede verse, respalda los resultados de las curvas de temperatura recién mencionadas, ya que la temperatura aumenta con el radio y el tiempo. Sin embargo, el resultado no es tan lineal como inicialmente parece, ya que si se examina $u(r,10)$ se obtiene la curva del régimen permanente, graficado en la figura \ref{figure:graphD}. Cabe mencionar que la curva exhibe una geometría levemente convexa. \\

Por último, se calcula la deformación del cilindro en función del tiempo. Dicha deformación puede verse en la figura \ref{figure:deformation}, la cual denota un comportamiento aproximadamente lineal. Como tanto la deformación como la temperatura son crecientes en el tiempo, se puede deducir que la deformación es linealmente proporcional a la temperatura. Además, el valor de la deformación para el régimen permanente es $d(10) = 1242.5$, lo cual se confirma con el gráfico de la figura \ref{figure:deformation}. \\


\begin{figure}[hbp]
	\centering
	\includegraphics[scale = 0.50]{img/graph_c.png}
	\caption{Temperatura del cilindro para $r = 0.6,0.7,0.8$ y $0.9$}
	\label{figure:graphC}
\end{figure}
\begin{figure}
	\centering
	\includegraphics[scale = 0.25]{img/graph_mesh.png}
	\caption{Evolución de la temperatura del cilindro}
	\label{figure:mesh}
\end{figure}
\begin{figure}
	\centering
	\includegraphics[scale = 0.35]{img/graph_d.png}
	\caption{Temperatura en régimen permamente}
	\label{figure:graphD}
\end{figure}
\begin{figure}
	\centering
	\includegraphics[scale = 0.35]{img/graph_e.png}
	\caption{Deformación del cilindro}
	\label{figure:deformation}
\end{figure}



\vspace{1cm}
\section{Conclusiones}
\label{section:conclusions}
\vspace{0.5cm}

Concluyendo, se exponen métodos numéricos para la resolución de ecuaciones diferenciales a derivadas parciales y se analizan los resultados obtenidos, graficándolos además en determinados casos. \\

Las temperaturas obtenidas evidencian una tendencia lineal y creciente en función de $r$ y $t$. Dicha tendencia puede ser útil para generar una aproximación lineal, reduciéndose así las complejidades temporales del cálculo. A su vez, la deformación es mayor a medida que la temperatura aumenta.



%-------------------------------------------------------------------------

\vspace{1cm}
\begin{thebibliography}{1}
\bibitem[Mathews y Fink(1992)]{mathews}
	Mathews, John H.,
	Fink, Kurtis D.,
	``Numerical Methods Using MATLAB'',
	Prentice Hall,
	1999
	
\end{thebibliography}



%-------------------------------------------------------------------------

\begin{figure}
	\lstinputlisting[label=lst:finiteDifferences,caption=Implementación del cálculo de $v_{k}^{m}$.]{../src/finiteDifferences.m}
\end{figure}

\begin{figure}
	\lstinputlisting[label=lst:stableDef,caption=Implementación del calculo de la deformación para un t.]{../src/stable_deformation.m}
\end{figure}

\begin{figure}
	\lstinputlisting[label=lst:graphe,caption=Implementación del cálculo de la deformación para $0 < t < 10$.]{../src/graph_e.m}
\end{figure}

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